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zmey [24]
3 years ago
14

Which pair shows equivalent expressions?

Mathematics
1 answer:
Simora [160]3 years ago
7 0

Which pair shows equivalent expressions?

A.2(2/5x + 2)=2 2/5x + 1

B.2(2/5x + 2)=4/5x + 4

C.2(2/5x + 4)=4/5x + 2

D.2(2/5x + 4)=2 2/5x + 8

Solution:

Let us distribute 2 inside the parenthesis.

That is, we use distributive property:

a(b+c)=ab+ac

So,

Answer:Option (b)

Applying distributive property, a(b+c)=ab+ac

So, Option (B) is correct.

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How many integers between 10 and 60, inclusive, can be evenly divided by neither 2 nor 3?
erastovalidia [21]
Find total number of integers.

a_1=10,a_n=60,d=1 \\a_n=a_1+(n-1)d \\60=10+(n-1)1 \\n-1=50 \\n=51

Find how many integers is divisible by 2.

a_1=10,a_n=60,d=2&#10;\\a_n=a_1+(n-1)d&#10;\\60=10+(n-1)2&#10;\\2(n-1)=50&#10;\\n-1=25&#10;\\n=26

Eliminate even numbers.

11, 13, 15,..., 57, 59

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Eliminate numbers before the first number divisible by 3 and after the last number divisible by 3.

15, 17, 19,..., 55, 57

This array contains 25 - 3 = 22 numbers.

Now we should eliminate numbers divisible by 3: 15, 21, 27...

a_1=15,a_n=57,d=6,n=?&#10;\\a_n=a_1+(n-1)d&#10;\\57=15+(n-1)6&#10;\\6(n-1)=42&#10;\\n-1=7&#10;\\n=8

There are 8 such numbers.

Therefore, there are 25 - 8 = 17 numbers that <span>can be evenly divided by neither 2 nor 3</span>

6 0
3 years ago
Need help ASAP!!! <br> Multiple choice answers
kobusy [5.1K]

Answer:

A. 16.55

Step-by-step explanation:

What we know is the angle of 38° and the hypotenuse is 21. Before we find x (side b), we have to find side a first. TO find that multiply the sine of 38 by 21.

21 * sin(38)

<em>sin(38) calculates to 0.615661475.</em>

21 * 0.615661475

<em>Multiply 21 by 0.615661475</em>

side a = 12.928890975

Now to find x. Use this to find x: √21² - 12.928890975²

√21² - 12.928890975²

<em>Square both numbers</em>

√441 - 167.156221843

<em>Subtract 167.156221843 from 441</em>

√273.843778157

<em>Find the square root of 273.843778157</em>

x ≈ 16.548225831

x ≈ 16.55, so the answer is A.

8 0
3 years ago
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